help me with this math problem before I go mad

[Random Variable]

Can you write x^3-x+e^x=0 as x=g(x) such that the absolute value of the derivative of g(x) evaluated at approximinately x=-1.1 is less than 1 ?

The most obvious choices x=x^3+e^x, x=ln(x-x^3), and x=(x-e^x)^(1/3) all don't work. A lot of other rearrangements don't work either.

EDIT: x= -e^x/x^2+1/x doesn't work

SECOND EDIT: I found one that works! x=(3x-e^x-x^3)/2

 

[BrunoPuntzJones]

<(^^)>

 

[RapidSnail]

Originally posted by: BrunoPuntzJones
<(^^)>

t(^.^t)

 

[WildHorse]

 

[dullard]

Misread your post.

 

[Random Variable]

Originally posted by: scott
1) numerical solution is:

-1 + 2.71828^(1+y) -y+(1+y)^2 = 0

2) now evaluate your derivative on that

(Not to spoil your fun in working it out, for x=-1.1 it evaluates to "false")

The root is not exactly -1.1.

 

[Syringer]

Originally posted by: Random Variable
Can you write x^3-x+e^x=0 as x=g(x) such that the absolute value of the derivative of g(x) evaluated at approximinately x=-1.1 is less than 1 ?

The answer is 'no'.

What do I win?